MCQ
$\text{If}\ \sin\alpha+\sin\beta=\text{a}\text{ and }\cos\alpha-\cos\beta=\text{b},\ \text{than }\tan\frac{\alpha-\beta}{2}=$
- A$-\frac{\text{a}}{\text{b}}$
- B$-\frac{\text{b}}{\text{a}}$
- C$\sqrt{\text{a}^2+\text{b}^2}$
- DNone of these
Solution:$$
Given:
$\sin\alpha+\sin\beta=\text{a}...(\text{i})$
$\cos\alpha-\cos\beta=\text{b}...(\text{ii})$
Dividing (i) by (ii):
$\Rightarrow\ \frac{\sin\alpha+\sin\beta}{\cos\alpha-\cos\beta}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \frac{2\sin\big(\frac{\alpha+\beta}{2}\big)\cos\big(\frac{\alpha-\beta}{2}\big)}{-2\sin\big(\frac{\alpha+\beta}{2}\big)\sin\big(\frac{\alpha-\beta}{2}\big)}=\frac{\text{a}}{\text{b}}$ $\Big[\because\ \sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\text{ and }\cos\text{A}+\cos\text{B}$
$\Rightarrow\ \frac{\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)}{-\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \cot\Big(\frac{\alpha-\beta}{2}\Big)=-\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \frac{1}{\cot\Big(\frac{\alpha-\beta}{2}\Big)}=\frac{1}{-\frac{\text{a}}{\text{b}}}$
$\Rightarrow\ \tan\Big(\frac{\alpha-\beta}{2}\Big)=-\frac{\text{b}}{\text{a}}$
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